Problem: Graph this system of equations and solve. $y = \dfrac{1}{3} x - 1$ $8x+6y = 24$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $\dfrac{1}{3}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $3$ positions to the right. $3$ positions to the right. Graph the blue line so it passes through $(0, -1)$ and $(3, 0)$ Convert the second equation, $8x+6y = 24$ , to slope-intercept form. $y = -\dfrac{4}{3} x + 4$ The y-intercept for the second equation is $4$ , so the second line must pass through the point $(0, 4)$ The slope for the second equation is $-\dfrac{4}{3}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. $4$ positions down from $(0, 4)$ is $(3, 0)$ Graph the green line so it passes through $(0, 4)$ and $(3, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(3, 0)$.